3.483 \(\int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac{4 \cos (c+d x)}{a^2 d \sqrt{a \sin (c+d x)+a}}-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}+\frac{4 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

(-4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) + (4*Cos[c + d*x]^
5)/(7*d*(a + a*Sin[c + d*x])^(5/2)) + (2*Cos[c + d*x]^3)/(3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (2*Cos[c + d*x]^
5)/(7*a*d*(a + a*Sin[c + d*x])^(3/2)) + (4*Cos[c + d*x])/(a^2*d*Sqrt[a + a*Sin[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.428343, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2878, 2860, 2679, 2649, 206} \[ \frac{4 \cos (c+d x)}{a^2 d \sqrt{a \sin (c+d x)+a}}-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{2 \cos ^5(c+d x)}{7 a d (a \sin (c+d x)+a)^{3/2}}+\frac{4 \cos ^5(c+d x)}{7 d (a \sin (c+d x)+a)^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) + (4*Cos[c + d*x]^
5)/(7*d*(a + a*Sin[c + d*x])^(5/2)) + (2*Cos[c + d*x]^3)/(3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (2*Cos[c + d*x]^
5)/(7*a*d*(a + a*Sin[c + d*x])^(3/2)) + (4*Cos[c + d*x])/(a^2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2878

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*g*(m + p + 2)), x] + Dist[1/
(b*(m + p + 2)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*(p + 1)*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 2, 0]

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac{2 \int \frac{\cos ^4(c+d x) \left (-\frac{3 a}{2}-5 a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac{4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\int \frac{\cos ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\\ &=\frac{4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac{2 \int \frac{\cos ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{a}\\ &=\frac{4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x)}{a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{4 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac{4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x)}{a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{8 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^2 d}\\ &=-\frac{4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac{4 \cos ^5(c+d x)}{7 d (a+a \sin (c+d x))^{5/2}}+\frac{2 \cos ^3(c+d x)}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^5(c+d x)}{7 a d (a+a \sin (c+d x))^{3/2}}+\frac{4 \cos (c+d x)}{a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.30096, size = 201, normalized size = 1.19 \[ -\frac{\sqrt{a (\sin (c+d x)+1)} \left (525 \sin \left (\frac{1}{2} (c+d x)\right )+91 \sin \left (\frac{3}{2} (c+d x)\right )-21 \sin \left (\frac{5}{2} (c+d x)\right )-3 \sin \left (\frac{7}{2} (c+d x)\right )-525 \cos \left (\frac{1}{2} (c+d x)\right )+91 \cos \left (\frac{3}{2} (c+d x)\right )+21 \cos \left (\frac{5}{2} (c+d x)\right )-3 \cos \left (\frac{7}{2} (c+d x)\right )+(672+672 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \sec \left (\frac{d x}{4}\right ) \left (\cos \left (\frac{1}{4} (2 c+d x)\right )-\sin \left (\frac{1}{4} (2 c+d x)\right )\right )\right )\right )}{84 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-(Sqrt[a*(1 + Sin[c + d*x])]*((672 + 672*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c +
d*x)/4] - Sin[(2*c + d*x)/4])] - 525*Cos[(c + d*x)/2] + 91*Cos[(3*(c + d*x))/2] + 21*Cos[(5*(c + d*x))/2] - 3
*Cos[(7*(c + d*x))/2] + 525*Sin[(c + d*x)/2] + 91*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] - 3*Sin[(7*(c
+ d*x))/2]))/(84*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 0.878, size = 132, normalized size = 0.8 \begin{align*} -{\frac{2+2\,\sin \left ( dx+c \right ) }{21\,{a}^{6}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 42\,{a}^{7/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) -3\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{7/2}-7\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}-42\,{a}^{3}\sqrt{a-a\sin \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-2/21*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(42*a^(7/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/
a^(1/2))-3*(a-a*sin(d*x+c))^(7/2)-7*(a-a*sin(d*x+c))^(3/2)*a^2-42*a^3*(a-a*sin(d*x+c))^(1/2))/a^6/cos(d*x+c)/(
a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^2/(a*sin(d*x + c) + a)^(5/2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.17619, size = 711, normalized size = 4.21 \begin{align*} \frac{2 \,{\left (\frac{21 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac{2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt{a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt{a}} +{\left (3 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{3} - 31 \, \cos \left (d x + c\right )^{2} +{\left (3 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right ) - 80\right )} \sin \left (d x + c\right ) + 61 \, \cos \left (d x + c\right ) + 80\right )} \sqrt{a \sin \left (d x + c\right ) + a}\right )}}{21 \,{\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/21*(21*sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c)
- 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x
+ c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) + (3*cos(d*x + c)^4 - 9*cos(d*x + c)^3 -
31*cos(d*x + c)^2 + (3*cos(d*x + c)^3 + 12*cos(d*x + c)^2 - 19*cos(d*x + c) - 80)*sin(d*x + c) + 61*cos(d*x +
c) + 80)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.47393, size = 486, normalized size = 2.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/1344*(10752*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) +
sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - (((((((13*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan
(1/2*d*x + 1/2*c)/a^11 - 21*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 56*sgn(tan(1/2*d*x + 1/
2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 70*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) + 70*sgn(ta
n(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 56*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*
c) + 21*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)*tan(1/2*d*x + 1/2*c) - 13*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^11)/(a*t
an(1/2*d*x + 1/2*c)^2 + a)^(7/2) - 4*sqrt(2)*(2688*a^(27/2)*arctan(sqrt(a)/sqrt(-a)) + 5*sqrt(-a)*a)*sgn(tan(1
/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^(31/2)))/d